If a cell has a standard electrode potential of $0.295 \ V$ and $n = 2$,calculate its equilibrium constant at $298 \ K$.

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.
$(1 \ F = 96,500 \ C \ mol^{-1})$

Consider the following electrochemical cell at $298 \ K$:
$Pt | HSnO_2^-(aq) | Sn(OH)_6^{2-}(aq) || Bi_2O_3(s) | Bi(s)$.
If the reaction quotient at a given time is $10^6$, then the cell $EMF$ $(E_{\text{cell}})$ is . . . . . . $\times 10^{-1} \ V$ (Nearest integer).
Given the standard half-cell reduction potential as
$E^0_{Bi_2O_3/Bi, OH^-} = -0.44 \ V$ and
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.

For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

Calculate the equilibrium constant for the following reaction: $Ni_{(s)} + Cu_{(aq)}^{2+} \to Cu_{(s)} + Ni_{(aq)}^{2+}$. Given: $E_{Ni^{2+}|Ni}^o = -0.25 \ V$ and $E_{Cu^{2+}|Cu}^o = 0.34 \ V$.

The standard electrode potential of a $Cu^{2+} | Cu$ electrode is $0.34 \, V$ (reduction potential). What will be the electrode potential of a $0.001 \, M \, Cu^{2+}$ solution in $V$?